Key Concepts and Formulas

• Equation of a Circle: The general equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Its center is $( -g, -f)$, and its radius is $r = \sqrt {{g^2} + {f^2} - c}$.
• Angle in a Semicircle: An angle inscribed in a semicircle is a right angle. Conversely, the angle subtended by a diameter at any point on the circumference is a right angle.
• Pythagorean Theorem: In a right-angled triangle, ${a^2} + {b^2} = {c^2}$, where c is the hypotenuse.
• Area of a Triangle: Area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

Step-by-Step Solution

Step 1: Determine the Center and Radius of the Given Circle

The given equation of the circle is ${x^2} - \sqrt 2 (x + y) + {y^2} = 0$. We need to rewrite it in the standard form ${x^2} + {y^2} + 2gx + 2fy + c = 0$ to find its center and radius.

First, distribute the $-\sqrt{2}$: ${x^2} + {y^2} - \sqrt 2 x - \sqrt 2 y = 0$

Now, compare this with the general form. We can identify the coefficients:

• $2g = -\sqrt{2} \Rightarrow g = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$
• $2f = -\sqrt{2} \Rightarrow f = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$
• $c = 0$

Now, let's find the center and radius:

• Center: $( -g, -f) = \left( { - \left( { - \frac{1}{{\sqrt 2 }}} \right), - \left( { - \frac{1}{{\sqrt 2 }}} \right)} \right) = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$
• Radius: $r = \sqrt {{g^2} + {f^2} - c}$. Substituting the values of $g$, $f$, and $c$: $r = \sqrt {{{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2} - 0}$ $r = \sqrt {\frac{1}{2} + \frac{1}{2} - 0}$ $r = \sqrt 1$ $r = 1$ Explanation: The radius of the circle is 1. This is important because the hypotenuse of the right-angled triangle is the diameter, which is twice the radius.

Step 2: Relate the Triangle's Angle to the Circle's Diameter

The problem states that the triangle ABC is inscribed in the circle and $\angle BAC = {\pi \over 2} = 90^\circ$. Because $\angle BAC$ is a right angle and the triangle is inscribed in the circle, the side opposite the right angle, BC, must be the diameter of the circle.

Therefore, the length of side BC is equal to the diameter of the circle. Length of BC = $2 \times \text{radius} = 2 \times r$. Since $r = 1$, Length of BC $= 2 \times 1 = 2$ units.

Explanation: We used the "angle in a semicircle" property to determine that BC is the diameter and has a length of 2.

Step 3: Calculate the Length of the Missing Side AC using the Pythagorean Theorem

We now know that $\triangle ABC$ is a right-angled triangle with the right angle at A. We have:

• Hypotenuse BC = 2 units (calculated in Step 2).
• One leg AB = $\sqrt{2}$ units (given in the problem).
• The other leg AC = ?

Using the Pythagorean theorem, ${AB^2} + {AC^2} = {BC^2}$: ${(\sqrt{2})^2} + {AC^2} = {(2)^2}$ $2 + {AC^2} = 4$ ${AC^2} = 4 - 2$ ${AC^2} = 2$ $AC = \sqrt{2}$ units (since length must be positive).

Explanation: We used the Pythagorean theorem and the lengths of AB and BC to find the length of AC.

Step 4: Calculate the Area of $\triangle ABC$

The area of a right-angled triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$. In $\triangle ABC$, the legs AB and AC can be considered as the base and height. Area of $\triangle ABC = \frac{1}{2} \times AB \times AC$.

Substituting the values: Area of $\triangle ABC = \frac{1}{2} \times \sqrt{2} \times \sqrt{2}$ Area of $\triangle ABC = \frac{1}{2} \times 2$ Area of $\triangle ABC = 1$ square unit.

Explanation: We calculated the area of the triangle using the lengths of the two legs, which we now know.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when finding the center of the circle. It's $( -g, -f)$, not $(g, f)$.
• Confusing Radius and Diameter: Remember that the diameter is twice the radius. A common mistake is to equate the side opposite the right angle to the radius instead of the diameter.
• Pythagorean Theorem Mix-up: Ensure you correctly identify the hypotenuse before applying the Pythagorean theorem.

Summary

This problem required us to find the radius and center of a circle from its general equation, apply the "angle in a semicircle" theorem, use the Pythagorean theorem, and finally calculate the area of a right-angled triangle. The key was to recognize that the hypotenuse of the triangle was the diameter of the circle.

The final answer is 1.

Final Answer

The final answer is \boxed{1}, which corresponds to option (A).